HW #2 Computer Architecture
Due: 9/17/04 (F)
1. Compare zero-, one-, two-, three-address, and the load & store machines by writing programs to compute
X = A + B * C - E;
Y = A / (E * C - D);
for each of the five machines. The instructions available for use are as follows:
| 3 Address | 2 Address | 1 Address (Accumulator machine) |
0 Address (Stack machine) |
MOVE (X  Y) |
MOVE (X  Y) |
LOAD M | PUSH M |
| STORE M | POP M | ||
ADD (X  Y + Z) |
ADD (X  X + Y) |
ADD M | ADD |
SUB (X  Y - Z) |
ADD (X  X - Y) |
SUB M | SUB |
MUL (X  Y * Z) |
MUL (X  X * Y) |
MUL M | MUL |
DIV (X  Y / Z) |
DIV (X  X / Y) |
DIV M | DIV |
| Notes: "SUB M" performs AC = AC - M "DIV M" performs AC = AC / M |
Notes: "SUB" performs POP T POP T2 T3 = T2 - T PUSH T3 "DIV" performs POP T POP T2 T3 = T2 / T PUSH T3 |
Load/Store Architecture - operands for arithmetic operations must be from/to registers. For example, to perform the high-level statement "Z = X - Y" we need the code:
LOAD R2, X
LOAD R3, Y
SUB R4, R2, R3
STORE R4, Z
| 3 Address | 2 Address | 1 Address | 0 Address | Load & Store |
| MUL X, B, C ADD X, A, X SUB X, X, E MUL Y, E, C SUB Y, Y, D DIV Y, A, Y |
MOVE X, B MUL X, C ADD X, A SUB X, E MOVE T, E MUL T, C SUB T, D MOVE Y, A DIV Y, T |
LOAD B MUL C ADD A SUB E STORE X LOAD E MUL C SUB D STORE Y LOAD A DIV Y STORE Y |
PUSH A PUSH B PUSH C MUL ADD PUSH E SUB POP X PUSH A PUSH E PUSH C MUL PUSH D SUB DIV POP Y |
LOAD R1, B LOAD R2, C MUL R3, R1, R2 LOAD R1, A ADD R3, R3, R1 LOAD R4, E SUB R3, R3, R4 STORE R3, X MUL R3, R4, R2 LOAD R4, D SUB R3, R3, R4 DIV R3, R1, R3 STORE R3, Y |
2. Assume 8-bit opcodes, 32-bit absolute addressing, 5-bit register numbers, and 32-bit operands. Compute the number of bits needed in programs from question 1 by completing the following table.
| 3 Address | 2 Address | 1 Address | 0 Address | Load & Store | |
| Number of bits needed to store the program | 6 x (8+3x32) = 624 | 9 x (8+2x32) = 648 | 12x(8+32) = 480 | 10x(8+32) + 6x8 = 448 |
7x(8+32+5) + 6x(8+5+5+5)= 453 |
| Number of bits of data transferred to and from memory | 6x3x32 = 576 | 3x2x32 + 6x3x32 = 768 | 12x32 = 384 | 10x32 = 320 | 7 x 32 = 224 |
| Total number of bits read and written while the program executes | 1,200 | 1,416 | 864 | 768 | 677 |
3.
| Register File | Memory | |||
| R0: | (Label X:) 0 | 4 | ||
| R1: | 20 | 4 | 5 | |
| R2: | 4 | 8 | 6 | |
| R3: | 8 | 12 | 7 | |
| R4: | 12 | 16 | 8 | |
| R5: | 16 | 20 | 9 | |
| R6: | 20 | 24 | 10 | |
| 28 | 11 | |||
| 32 | 12 | |||
| 36 | 13 | |||
| PC: | 36 | 40 | 14 |
For each AL statement listed below, determine the value loaded into register R0. Always assume the above unmodified register and memory values.
| Assembly Language Statement | Type of addressing | R0's Register Value After Execution |
| Load R0, X | Direct | 4 |
| Load R0, (R1) | Register-Indirect | 9 |
| Load R0, 8(R1) | Base-register/Displacement | 11 |
| Load R0, -8(R1) | Base-register/Displacement | 7 |
| Load R0, X(R3) | Indexing | 6 if you assume R3 contains a byte displacement or 12 if you assume R3 contains an index into an array |